The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Practice Midterm — Chapters 21–28

Problems with worked solutions · 100 points · 5 extra credit

Practice Midterm — Electricity & Magnetism (Chapters 21–28)

Practice examination · 100 points core + 5 points extra credit · suggested time 2 hours. This practice exam is a parallel form of the real midterm: same chapters, same structure, same difficulty — only the numbers differ. Show all reasoning; partial credit is given. A calculator is allowed.

Constants: $k=8.99\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$, $\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$, $e=1.60\times10^{-19}\,\text{C}$, $m_e=9.11\times10^{-31}\,\text{kg}$, $m_p=1.67\times10^{-27}\,\text{kg}$, $\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}$.

1. 1. Electric Charge & the Electric Field (12 points). A charge $q_1=+5.0\,\text{nC}$ sits at the origin and a charge $q_2=-8.0\,\text{nC}$ sits at $(6.0\,\text{cm},\,0)$. Let $P$ be the point $(0,\,8.0\,\text{cm})$.

   (a) (3 pts) Find the magnitude of the electrostatic force between $q_1$ and $q_2$, and state whether it is attractive or repulsive.

   (b) (6 pts) Find the net electric field at $P$: give its magnitude and its direction (as an angle measured from the $+x$ axis).

   (c) (3 pts) A charge $q_3=-3.0\,\text{nC}$ is placed at $P$. Find the magnitude and direction of the force on it.

   Solution

   1.

   (a) Opposite signs, so the force is attractive: $F=\dfrac{k|q_1q_2|}{r^2}=\dfrac{(8.99\times10^{9})(5.0\times10^{-9})(8.0\times10^{-9})}{(0.060)^2}\approx 1.0\times10^{-4}\,\text{N}$.

   (b) $q_1$ is $8.0\,\text{cm}$ from $P$ (along $+y$); $q_2$ is $\sqrt{6^2+8^2}=10\,\text{cm}$ from $P$. The field of $q_1$ (positive) points away from it, i.e. $+y$: $E_1=\dfrac{k(5.0\times10^{-9})}{(0.080)^2}\approx 7.0\times10^{3}\,\text{N/C}$. The field of $q_2$ (negative) points toward $q_2$, along the unit vector $(6,-8)/10=(0.6,-0.8)$: $E_2=\dfrac{k(8.0\times10^{-9})}{(0.10)^2}\approx 7.2\times10^{3}\,\text{N/C}$.

   Components: $E_x=E_2(0.6)\approx 4.3\times10^{3}$, $E_y=E_1+E_2(-0.8)\approx 7.0\times10^{3}-5.8\times10^{3}=1.3\times10^{3}\,\text{N/C}$. Magnitude $|\mathbf{E}|=\sqrt{E_x^2+E_y^2}\approx 4.5\times10^{3}\,\text{N/C}$, at $\theta=\tan^{-1}(1.27/4.32)\approx 16^\circ$ above the $+x$ axis.

   (c) $F=|q_3|E=(3.0\times10^{-9})(4.5\times10^{3})\approx 1.4\times10^{-5}\,\text{N}$. Because $q_3$ is negative, the force is opposite to $\mathbf{E}$: directed $16^\circ$ below the $-x$ axis.

2. 2. Gauss's Law (12 points). A point charge $q=+3.0\,\text{nC}$ sits at the center of a neutral conducting spherical shell of inner radius $a=5.0\,\text{cm}$ and outer radius $b=8.0\,\text{cm}$.

   (a) (3 pts) Find the charge induced on the inner surface and on the outer surface of the shell.

   (b) (6 pts) Find the magnitude of the electric field at $r=3.0\,\text{cm}$, at $r=6.5\,\text{cm}$, and at $r=12\,\text{cm}$.

   (c) (3 pts) Find the net electric flux through a spherical Gaussian surface of radius $r=6.5\,\text{cm}$, and explain in one sentence why it has that value.

   Solution

   2.

   (a) The field must vanish inside the conductor, so the inner surface carries $-3.0\,\text{nC}$ (canceling the enclosed charge). The shell is neutral, so the outer surface carries $+3.0\,\text{nC}$.

   (b) $r=3.0\,\text{cm}$ (inside the cavity): only the central charge acts, $E=\dfrac{kq}{r^2}=\dfrac{(8.99\times10^{9})(3.0\times10^{-9})}{(0.030)^2}\approx 3.0\times10^{4}\,\text{N/C}$. $r=6.5\,\text{cm}$ (within the conductor): $E=0$. $r=12\,\text{cm}$ (outside): $E=\dfrac{kq}{r^2}=\dfrac{(8.99\times10^{9})(3.0\times10^{-9})}{(0.12)^2}\approx 1.9\times10^{3}\,\text{N/C}$.

   (c) $\Phi=0$. The Gaussian surface at $r=6.5\,\text{cm}$ lies inside the conductor and encloses the central $+3.0\,\text{nC}$ plus the induced $-3.0\,\text{nC}$ on the inner surface, a net enclosed charge of zero.

3. 3. Electric Potential (13 points). Take $V=0$ at infinity throughout.

   (a) (4 pts) A charge $q_1=+4.0\,\text{nC}$ is at the origin and $q_2=-3.0\,\text{nC}$ is at $(0.40\,\text{m},\,0)$. Find the potential at the midpoint $P=(0.20\,\text{m},\,0)$, and the work required to bring a $+2.0\,\text{nC}$ charge from infinity to $P$.

   (b) (4 pts) Two large parallel plates are $8.0\,\text{mm}$ apart with a potential difference of $300\,\text{V}$ between them. Find the electric field magnitude between the plates, and the speed an electron gains when it starts from rest at the negative plate and reaches the positive plate.

   (c) (5 pts) In a region of space the potential is $V(x,y,z)=2x^2y-3yz+z^2$ (in volts, with $x,y,z$ in metres). Find $\mathbf{E}(x,y,z)$, evaluate it at the point $(1,\,2,\,-1)\,\text{m}$, and give its magnitude there.

   Solution

   3.

   (a) Both charges are $0.20\,\text{m}$ from $P$: $V=\dfrac{k(q_1+q_2)}{r}=\dfrac{(8.99\times10^{9})(1.0\times10^{-9})}{0.20}\approx 45\,\text{V}$. Work: $W=qV=(2.0\times10^{-9})(45)\approx 9.0\times10^{-8}\,\text{J}$.

   (b) $E=\dfrac{V}{d}=\dfrac{300}{8.0\times10^{-3}}=3.75\times10^{4}\,\text{V/m}$. Energy conservation $\tfrac12 m_e v^2=eV$: $v=\sqrt{\dfrac{2eV}{m_e}}=\sqrt{\dfrac{2(1.60\times10^{-19})(300)}{9.11\times10^{-31}}}\approx 1.0\times10^{7}\,\text{m/s}$.

   (c) $\mathbf{E}=-\nabla V=\left(-4xy,\;-(2x^2-3z),\;-(2z-3y)\right)=(-4xy,\;-2x^2+3z,\;3y-2z)\,\text{V/m}$. At $(1,2,-1)$: $\mathbf{E}=(-8,\,-5,\,8)\,\text{V/m}$, with $|\mathbf{E}|=\sqrt{8^2+5^2+8^2}=\sqrt{153}\approx 12\,\text{V/m}$.

4. 4. Capacitance & Dielectrics (13 points).

   (a) (4 pts) A parallel-plate capacitor has plate area $A=0.015\,\text{m}^2$ and separation $d=2.0\,\text{mm}$, connected to a $24\,\text{V}$ battery. Find the capacitance, the charge on each plate, the energy stored, and the electric field between the plates.

   (b) (5 pts) A capacitor $C_1=6.0\,\mu\text{F}$ is connected in series with the parallel combination of $C_2=4.0\,\mu\text{F}$ and $C_3=2.0\,\mu\text{F}$; the network is placed across an $18\,\text{V}$ battery. Find the equivalent capacitance, the total charge delivered, and the voltage across and charge on each capacitor.

   (c) (4 pts) A $5.0\,\mu\text{F}$ capacitor is charged to $40\,\text{V}$ and then disconnected from the battery. Its plate separation is then doubled. Find the new capacitance, charge, and voltage, and the change in stored energy — and state where that energy came from.

   Solution

   4.

   (a) $C=\dfrac{\varepsilon_0 A}{d}=\dfrac{(8.85\times10^{-12})(0.015)}{2.0\times10^{-3}}\approx 6.6\times10^{-11}\,\text{F}=66\,\text{pF}$. $Q=CV\approx(6.64\times10^{-11})(24)\approx 1.6\times10^{-9}\,\text{C}$. $U=\tfrac12 CV^2\approx 1.9\times10^{-8}\,\text{J}$. $E=V/d=24/(2.0\times10^{-3})=1.2\times10^{4}\,\text{V/m}$.

   (b) Parallel pair: $C_2+C_3=6.0\,\mu\text{F}$. In series with $C_1$: $C_\text{eq}=\dfrac{(6)(6)}{6+6}=3.0\,\mu\text{F}$. Total charge $Q=C_\text{eq}V=(3.0)(18)=54\,\mu\text{C}$. $C_1$ (in series) carries the full $54\,\mu\text{C}$, so $V_1=54/6=9.0\,\text{V}$; the remaining $9.0\,\text{V}$ is across the pair, giving $Q_2=(4.0)(9.0)=36\,\mu\text{C}$ and $Q_3=(2.0)(9.0)=18\,\mu\text{C}$ (which sum to $54\,\mu\text{C}$).

   (c) Doubling $d$ halves $C$: $C_\text{new}=2.5\,\mu\text{F}$. Disconnected, so $Q$ is fixed at $Q=(5.0\,\mu\text{F})(40)=200\,\mu\text{C}$. Then $V_\text{new}=Q/C_\text{new}=200/2.5=80\,\text{V}$. Energy: $U_i=\tfrac12(5.0\times10^{-6})(40)^2=4.0\times10^{-3}\,\text{J}$, $U_f=\tfrac12(2.5\times10^{-6})(80)^2=8.0\times10^{-3}\,\text{J}$, so $\Delta U=+4.0\times10^{-3}\,\text{J}$ — supplied by the work you do pulling the plates apart against their attraction.

5. 5. Current, Resistance & EMF (12 points).

   (a) (4 pts) A nichrome wire ($\rho=1.10\times10^{-6}\,\Omega\!\cdot\!\text{m}$) has length $2.0\,\text{m}$ and diameter $0.40\,\text{mm}$. Find its resistance. A second wire of the same material has twice the length and half the diameter; by what factor does its resistance change, and what is its resistance?

   (b) (5 pts) A battery of emf $12\,\text{V}$ and internal resistance $r=0.40\,\Omega$ drives an external resistor $R=5.6\,\Omega$. Find the current, the terminal voltage, and the power dissipated in $R$ and inside the battery; confirm these powers sum to the emf's output.

   (c) (3 pts) A space heater is rated $1500\,\text{W}$ at $120\,\text{V}$. Find its operating resistance and current. If it runs $2.0$ hours per day for $30$ days at $0.15 per kWh, find the energy used and the cost.

   Solution

   5.

   (a) $A=\pi(0.20\times10^{-3})^2\approx 1.26\times10^{-7}\,\text{m}^2$, so $R=\dfrac{\rho L}{A}=\dfrac{(1.10\times10^{-6})(2.0)}{1.26\times10^{-7}}\approx 17.5\,\Omega$. Since $R\propto L/d^2$, doubling $L$ and halving $d$ gives a factor $2\times4=8$: $R_\text{new}\approx 140\,\Omega$.

   (b) $I=\dfrac{\mathcal{E}}{R+r}=\dfrac{12}{5.6+0.40}=2.0\,\text{A}$. Terminal voltage $V=\mathcal{E}-Ir=12-(2.0)(0.40)=11.2\,\text{V}$. $P_R=I^2R=(2.0)^2(5.6)=22.4\,\text{W}$, $P_r=I^2r=(2.0)^2(0.40)=1.6\,\text{W}$; sum $=24\,\text{W}=\mathcal{E}I$. ✓

   (c) $R=\dfrac{V^2}{P}=\dfrac{120^2}{1500}=9.6\,\Omega$; $I=\dfrac{P}{V}=\dfrac{1500}{120}=12.5\,\text{A}$. Energy $=1.5\,\text{kW}\times(2.0\times30)\,\text{h}=90\,\text{kWh}$; cost $=90\times0.15=13.5$, i.e. $13.50.

6. 6. Direct-Current Circuits (13 points).

   (a) (6 pts) A $36\,\text{V}$ battery (ideal, no internal resistance) is connected to a resistor $R_1=4.0\,\Omega$ in series with the parallel combination of $R_2=6.0\,\Omega$ and $R_3=12\,\Omega$. Find the equivalent resistance, the total current, the voltage across the parallel combination, and the current through $R_2$ and through $R_3$.

   (b) (7 pts) An uncharged capacitor $C=25\,\mu\text{F}$ is connected in series with a resistor $R=200\,\text{k}\Omega$ and a $12\,\text{V}$ battery through a switch that closes at $t=0$. Find the time constant, the initial current, and the final charge. Then find the charge on the capacitor and the current in the circuit at $t=10\,\text{s}$, and the time required for the charge to reach $90\%$ of its final value.

   Solution

   6.

   (a) Parallel pair: $R_p=\dfrac{(6)(12)}{6+12}=4.0\,\Omega$. $R_\text{eq}=R_1+R_p=8.0\,\Omega$. $I=\dfrac{36}{8.0}=4.5\,\text{A}$. Voltage across the pair: $V_p=\mathcal{E}-IR_1=36-(4.5)(4.0)=18\,\text{V}$. Then $I_2=\dfrac{18}{6.0}=3.0\,\text{A}$ and $I_3=\dfrac{18}{12}=1.5\,\text{A}$ (sum $=4.5\,\text{A}$). ✓

   (b) $\tau=RC=(200\times10^{3})(25\times10^{-6})=5.0\,\text{s}$. Initial current $i(0)=\dfrac{\mathcal{E}}{R}=\dfrac{12}{200\times10^{3}}=60\,\mu\text{A}$. Final charge $Q_f=C\mathcal{E}=(25\times10^{-6})(12)=300\,\mu\text{C}$. At $t=10\,\text{s}$, $t/\tau=2$ and $e^{-2}=0.135$: $q=Q_f(1-e^{-2})=300(0.865)\approx 260\,\mu\text{C}$, and $i=i(0)e^{-2}=60(0.135)\approx 8.1\,\mu\text{A}$. For $90\%$: $1-e^{-t/\tau}=0.90\Rightarrow t=-\tau\ln(0.10)=5.0(2.303)\approx 12\,\text{s}$.

7. 7. Magnetic Field & Magnetic Forces (13 points).

   (a) (4 pts) A proton moves at $v=3.0\times10^{5}\,\text{m/s}$ at $30^\circ$ to a uniform field $B=0.60\,\text{T}$. Find the magnitude of the magnetic force. With $\mathbf{v}$ along $+x$ and $\mathbf{B}$ in the $xy$-plane (tilted $30^\circ$ above $+x$ toward $+y$), give the direction of the force using the right-hand rule, and state how it changes for an electron with the same velocity.

   (b) (5 pts) Starting from $\mathbf{F}=q\mathbf{v}\times\mathbf{B}$ and Newton's second law, derive the radius $r=mv/qB$ of the circular path of a particle moving perpendicular to a uniform field. Then, for an electron with $v=2.0\times10^{6}\,\text{m/s}$ perpendicular to $B=0.50\,\text{T}$, find the orbit radius, the period, and the cyclotron frequency.

   (c) (4 pts) A straight wire of length $L=0.40\,\text{m}$ carries a current $I=6.0\,\text{A}$ in a uniform field $B=0.50\,\text{T}$. Find the force on the wire when the current is perpendicular to $\mathbf{B}$, the force when the wire makes a $40^\circ$ angle with $\mathbf{B}$, and the angle at which the force is zero.

   Solution

   7.

   (a) $F=qvB\sin\theta=(1.60\times10^{-19})(3.0\times10^{5})(0.60)\sin30^\circ\approx 1.4\times10^{-14}\,\text{N}$. With $\mathbf{v}=v\hat{\mathbf{x}}$ and $\mathbf{B}$ in the $xy$-plane, $\mathbf{v}\times\mathbf{B}$ points along $+\hat{\mathbf{z}}$ (out of the page), and since $q>0$ the force is $+\hat{\mathbf{z}}$. For an electron the force reverses to $-\hat{\mathbf{z}}$ (into the page), same magnitude.

   (b) When $\mathbf{v}\perp\mathbf{B}$ the force $qvB$ is perpendicular to $\mathbf{v}$ and supplies the centripetal force: $qvB=\dfrac{mv^2}{r}\Rightarrow r=\dfrac{mv}{qB}$. For the electron, $r=\dfrac{(9.11\times10^{-31})(2.0\times10^{6})}{(1.60\times10^{-19})(0.50)}\approx 2.3\times10^{-5}\,\text{m}$. Period $T=\dfrac{2\pi m}{qB}\approx 7.2\times10^{-11}\,\text{s}$; cyclotron frequency $f=\dfrac{1}{T}=\dfrac{qB}{2\pi m}\approx 1.4\times10^{10}\,\text{Hz}$.

   (c) $F_\perp=BIL=(0.50)(6.0)(0.40)=1.2\,\text{N}$. At $40^\circ$: $F=BIL\sin40^\circ\approx 0.77\,\text{N}$. The force is zero when $\sin\theta=0$, i.e. at $\theta=0^\circ$ or $180^\circ$ (current parallel or antiparallel to $\mathbf{B}$).

8. 8. Sources of Magnetic Field (12 points).

   (a) (3 pts) A long straight wire carries $I=10\,\text{A}$. Find the magnetic field magnitude at a perpendicular distance $r=4.0\,\text{cm}$ from the wire, and the distance at which the field drops to $25\,\mu\text{T}$.

   (b) (4 pts) A flat circular coil has $N=40$ turns, radius $R=5.0\,\text{cm}$, and carries $I=3.0\,\text{A}$. Find the field magnitude at the center of the coil, and on the axis at $x=6.0\,\text{cm}$ from the center.

   (c) (5 pts) Two long straight parallel wires $3.0\,\text{cm}$ apart carry $I_1=12\,\text{A}$ and $I_2=18\,\text{A}$ in the same direction. Find the force per unit length between them and state whether they attract or repel. Separately, find the field inside an ideal solenoid that is $30\,\text{cm}$ long, wound with $900$ turns, carrying $2.5\,\text{A}$.

   Solution

   8.

   (a) $B=\dfrac{\mu_0 I}{2\pi r}=\dfrac{(4\pi\times10^{-7})(10)}{2\pi(0.040)}\approx 5.0\times10^{-5}\,\text{T}$. Setting $B=25\,\mu\text{T}$: $r=\dfrac{\mu_0 I}{2\pi B}=\dfrac{(4\pi\times10^{-7})(10)}{2\pi(25\times10^{-6})}=0.080\,\text{m}=8.0\,\text{cm}$.

   (b) Center: $B=\dfrac{\mu_0 N I}{2R}=\dfrac{(4\pi\times10^{-7})(40)(3.0)}{2(0.050)}\approx 1.5\times10^{-3}\,\text{T}$. On the axis: $B=\dfrac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}$ with $R^2+x^2=(0.050)^2+(0.060)^2=6.1\times10^{-3}\,\text{m}^2$, giving $B\approx 4.0\times10^{-4}\,\text{T}$.

   (c) $\dfrac{F}{L}=\dfrac{\mu_0 I_1 I_2}{2\pi d}=\dfrac{(4\pi\times10^{-7})(12)(18)}{2\pi(0.030)}\approx 1.4\times10^{-3}\,\text{N/m}$; the currents run the same way, so the wires attract. Solenoid: $n=\dfrac{900}{0.30}=3000\,\text{m}^{-1}$, so $B=\mu_0 n I=(4\pi\times10^{-7})(3000)(2.5)\approx 9.4\times10^{-3}\,\text{T}$.

9. Extra Credit (5 points)

   A solid insulating sphere of radius $R$ carries a charge density that increases linearly from the center, $\rho(r)=\rho_0\,\dfrac{r}{R}$ for $r\le R$ (and zero outside).

   (a) Find the total charge $Q$ of the sphere.

   (b) Use Gauss's law to find the electric field magnitude $E(r)$ both inside ($r<R$) and outside ($r>R$).

   (c) Show that your outside result reduces to the point-charge field $kQ/r^2$.

   Solution

   Extra Credit

   (a) $Q=\displaystyle\int_0^R \rho_0\frac{r}{R}\,4\pi r^2\,dr=\frac{4\pi\rho_0}{R}\int_0^R r^3\,dr=\frac{4\pi\rho_0}{R}\cdot\frac{R^4}{4}=\pi\rho_0 R^3.$

   (b) For $r<R$ the enclosed charge is $q_\text{enc}(r)=\dfrac{4\pi\rho_0}{R}\cdot\dfrac{r^4}{4}=\dfrac{\pi\rho_0 r^4}{R}$. Gauss's law $E(4\pi r^2)=q_\text{enc}/\varepsilon_0$ gives

   $$E(r)=\frac{\rho_0 r^2}{4\varepsilon_0 R}\quad(r<R).$$

   For $r>R$ all of $Q$ is enclosed: $E(r)=\dfrac{Q}{4\pi\varepsilon_0 r^2}=\dfrac{\rho_0 R^3}{4\varepsilon_0 r^2}.$

   (c) Using $Q=\pi\rho_0 R^3$, the outside field is $E=\dfrac{\rho_0 R^3}{4\varepsilon_0 r^2}=\dfrac{Q}{4\pi\varepsilon_0 r^2}=\dfrac{kQ}{r^2}$ — the point-charge result, as expected since a spherically symmetric charge looks like a point charge from outside.